Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{y^2 - 18y + 81}{y + 4} \times \dfrac{4y + 16}{3y - 27} $
Answer: First factor the quadratic. $n = \dfrac{(y - 9)(y - 9)}{y + 4} \times \dfrac{4y + 16}{3y - 27} $ Then factor out any other terms. $n = \dfrac{(y - 9)(y - 9)}{y + 4} \times \dfrac{4(y + 4)}{3(y - 9)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (y - 9)(y - 9) \times 4(y + 4) } { (y + 4) \times 3(y - 9) } $ $n = \dfrac{ 4(y - 9)(y - 9)(y + 4)}{ 3(y + 4)(y - 9)} $ Notice that $(y + 4)$ and $(y - 9)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ 4\cancel{(y - 9)}(y - 9)(y + 4)}{ 3(y + 4)\cancel{(y - 9)}} $ We are dividing by $y - 9$ , so $y - 9 \neq 0$ Therefore, $y \neq 9$ $n = \dfrac{ 4\cancel{(y - 9)}(y - 9)\cancel{(y + 4)}}{ 3\cancel{(y + 4)}\cancel{(y - 9)}} $ We are dividing by $y + 4$ , so $y + 4 \neq 0$ Therefore, $y \neq -4$ $n = \dfrac{4(y - 9)}{3} ; \space y \neq 9 ; \space y \neq -4 $